Question

a stone is projected vertically upwards from the ground with a velocity of 96 metre per second after 2 seconds another stone is projected upward from the same place with a velocity of 100 metre per second when and where the two stones meet each other​

Answer 1

Answer:

Fig not shown

so 37° angle not clear exactly

Assuming 37° with horizontal  & upward

Velocity = 10 m/s

Horizontal Velocity = Vcos37 = 10 * 0.8 = 8 m/s

Vertical Velocity = Vsin37 = 10* 0.6 = 6 m/s

Speed after 2 sec

Horizontal Velocity remains same = 8 m/s

Vertical Velocity = 6 - 10*2 = -14 m/s  => downward velocity

Velocity = √8² + 14² = 2√65  m/s  = 16.12 m/s

Max height attained when vertical velocity = 0

t = 6/10 = 0.6 sec

Vertical distance covered = 6²/(2 * 10) = 1.8 m

Max height from ground = 11 + 1.8 = 12.8 m

2 * 0.6 = 1.2 sec (time of flight for 11 m height)

Vertical velocity = 6 m/s down ward

11 m vertical distance will be covered in

6t + (1/2)10t² = 11

=> 5t² + 6t - 11 =0

=> 5t² -5t + 11t - 11 = 0

=> 5t(t - 1) + 11(t - 1) = 0

=> t = 1 sec

time of flight  1.2 + 1 = 2.2 sec

Horizontal Range = 8 * 2.2 = 17.6 m

Speed just before striking the ground

Vertical Velocity = 6 - 10*2.2 = 16 m/s

Horizontal = 8 m/s

Velocity = √8² + 16² = √320 = 8√5  =  17.89 m/s

Explanation: